Decimal representation

\(\frac{3}{28}\) is already in lowest terms, and \(28=2^2\times 7\). The decimal expansion of a rational number terminates only when the prime factors of its denominator are limited to \(2\) and \(5\). Since the denominator also contains the factor \(7\), the decimal expansion is non-terminating but recurring. The number is rational, so it is neither irrational nor an integer. Exam tip: After reducing a fraction, inspect its denominator; only factors \(2\) and \(5\) give a terminating decimal.

The fraction \(\frac{17}{125}\) is already in lowest terms, and \(125=5^3\). A fraction in lowest terms has a terminating decimal expansion when its denominator contains only the prime factors 2 and/or 5. In fact, \(\frac{17}{125}=\frac{136}{1000}=0.136\), so the decimal terminates. Option B is incorrect because the decimal is not infinite, and option C is incorrect because the fraction is rational. Exam tip: Check the prime factors of the denominator after reducing the fraction; only 2s and 5s give a terminating decimal.

Since \(1.69=1.3^2\), we have \(\sqrt{1.69}=1.3\). Similarly, \(0.09=0.3^2\), so \(\sqrt{0.09}=0.3\). Therefore, \(1.3+0.3=1.6\), making option A correct. Option D is only the value of the first square root and does not include the second term. Exam tip: express each decimal as the square of a simpler decimal number before finding its square root.

For a fraction in lowest terms, the decimal terminates only when the denominator has no prime factors other than 2 and 5. The fraction 8/13 is already in lowest terms, and 13 is a prime factor different from 2 and 5. Therefore its decimal expansion continues indefinitely but repeats in a cycle. Option B is correct; the fraction is defined and is not an integer.

The governing concept is the denominator test for rational decimals. A fraction in lowest terms has a terminating decimal expansion exactly when the prime factors of its denominator are only 2 and/or 5. Here, 200 = 2³ × 5², and 29 shares no factor with 200, so 29/200 is already in lowest terms. It can also be converted directly to denominator 1000: 29/200 = 145/1000 = 0.145. Since the decimal ends after three places, option A is correct. It is not irrational or non-terminating non-repeating; those descriptions apply to numbers such as √2. The expression is defined because its denominator is not zero.

A reduced fraction has a terminating decimal only if every prime factor of its denominator is 2 or 5. In 11/17, the denominator 17 is prime and is neither 2 nor 5. Long division therefore continues and eventually repeats a remainder, producing a non-terminating recurring decimal. Hence option B is correct; the fraction is defined but neither terminating nor an integer.

The governing criterion is that a rational number p/q in lowest terms has a terminating decimal expansion exactly when the prime factors of q are only 2 and/or 5. The fraction 33/160 is already in lowest terms because 33 and 160 have no common factor. Also, 160 = 2⁵ × 5, so the criterion is satisfied. To verify directly, multiply numerator and denominator by 625: 33/160 = 20625/100000 = 0.20625. The decimal ends, so option A is correct. It is not irrational, because it is a ratio of integers. It is not undefined because the denominator is non-zero, and it is not non-terminating non-repeating because its decimal expansion stops.

For a rational number written in lowest terms, the decimal expansion terminates only when the denominator has no prime factors other than 2 and 5. The fraction 7/66 is already in lowest terms, and 66 = 2 × 3 × 11. Since the denominator contains 3 and 11, its decimal expansion cannot terminate; every rational number nevertheless has either a terminating or a repeating decimal expansion. Therefore 7/66 is non-terminating but repeating, so option B is correct. The decimal begins 0.1060606..., showing a recurring pattern. Option A ignores the factors 3 and 11, option C confuses repeating rational decimals with irrational decimals, and option D is impossible because the fraction lies between 0 and 1.

A rational fraction in lowest terms has a terminating decimal expansion when the denominator contains only the prime factors 2 and 5. Here, 41 is prime and does not divide 500, so 41/500 is already reduced. Factorising the denominator gives 500 = 2² × 5³. Thus the criterion is satisfied and the decimal terminates. For a direct calculation, multiply by 2² to obtain 41/500 = 164/2000 = 0.082. Hence option A is correct. Options B and C describe a non-terminating non-repeating decimal, which cannot represent this rational fraction, while option D is impossible because the denominator is not zero.

The decimal-expansion rule must be applied after reducing the fraction. The greatest common divisor of 105 and 126 is 21, so 105/126 = 5/6. The denominator 6 factors as 2 × 3. Because the reduced denominator contains 3, not only 2 and 5, the decimal cannot terminate. Since 5/6 is rational, its decimal must be repeating; indeed, 5/6 = 0.83333..., where 3 repeats indefinitely. Hence option B is correct. Option A would be valid only if the reduced denominator had factors exclusively 2 and 5. Option C describes irrational decimals, and option D is impossible because the original denominator is non-zero.

The governing rule is that a rational number in lowest terms has a terminating decimal only when its denominator has no prime factors other than 2 and 5. First reduce the fraction: \(35/154=5/22\), because both numerator and denominator are divisible by 7. The denominator 22 factors as \(2\times11\), and the factor 11 remains. Therefore its decimal expansion does not terminate; because the number is rational, its decimal digits repeat periodically. In fact, \(5/22=0.2272727\ldots\). Hence option B is correct. It is not irrational, not terminating, and certainly not an integer.

rac{126}{224}=rac{9}{16} because both numerator and denominator are divisible by 14. In lowest form, the denominator is 16=2^4. A rational number rac{p}{q} has a terminating decimal expansion when, in lowest form, the prime factors of q are only 2 and/or 5. Hence, rac{9}{16}=0.5625 is terminating. A recurring decimal occurs when the denominator has a prime factor other than 2 or 5. Exam tip: reduce the fraction first, then factorise its denominator.

In option C, the digit 6 repeats indefinitely, so it is a recurring decimal. Every recurring decimal represents a rational number; for example, \(0.6666\ldots=\frac{2}{3}\). Options A and B have no fixed repeating block, and D is also not shown as a recurring decimal. Exam tip: Terminating and recurring decimals are always rational.

In the decimal 0.454545..., the block 45 repeats continuously, so it is a recurring decimal. Every recurring decimal is rational; here, 0.454545... = 45/99 = 5/11. It is neither an integer nor a natural number, while irrational numbers have non-terminating, non-recurring decimal expansions. Exam tip: A terminating or recurring decimal is always rational.

For \(\frac{13}{40}\), the denominator is \(40=2^3\times5\). A rational number in lowest terms has a terminating decimal expansion only when its denominator has no prime factors other than 2 and 5. \(\frac{7}{12}\) contains the factor 3 in its denominator, and \(\frac{11}{33}=\frac{1}{3}\), so they do not terminate. Exam tip: first reduce the fraction to lowest terms, then factorise the denominator.

In \(0.5000\ldots\), all digits after 5 are zeros, so its value remains \(0.5\). Now, \(0.5=\frac{5}{10}=\frac{1}{2}\); hence \(\frac{1}{2}\) is correct. The close distractor \(\frac{1}{5}\) equals \(0.2\), not \(0.5\). Exam tip: Zeros written at the end of a decimal do not change its value.

The decimal expansion of a rational number is either terminating or non-terminating repeating. Therefore, if its decimal expansion is infinite, a digit or a block of digits repeats, so option B is correct. A non-terminating non-repeating decimal represents an irrational number. Exam tip: For a rational number, an infinite decimal expansion must be repeating.

For \(\frac{9}{32}\), the denominator is \(32=2^5\). In lowest form, a rational number has a terminating decimal only when its denominator has prime factors \(2\) and/or \(5\) only. Hence, \(\frac{9}{32}=0.28125\) terminates. The denominators of \(\frac{5}{18}\) and \(\frac{11}{27}\) contain the factor \(3\), and \(\frac{7}{21}=\frac{1}{3}\) also has a factor \(3\) in its reduced denominator, so they are non-terminating recurring decimals. Exam tip: Reduce the fraction first, then factorise its denominator.

In 0.6666\ldots, the digit 6 repeats forever in a fixed pattern, so it is a recurring decimal. In 0.121221222\ldots and 0.1010010001\ldots, the lengths of the digit groups keep changing, while in 0.123456789\ldots the digits keep increasing; none has a fixed repeating block. Exam tip: in a recurring decimal, a digit or a fixed group of digits repeats indefinitely.

\(\frac{1}{3}=0.333\ldots\), where the digit 3 repeats forever. Hence, its decimal expansion is non-terminating recurring. In contrast, \(\frac{1}{8}=0.125\), \(\frac{5}{2}=2.5\), and \(\frac{7}{20}=0.35\) are terminating decimals. Exam tip: In lowest form, if a denominator has a prime factor other than 2 or 5, the decimal expansion is non-terminating recurring.